GATE Overflow | Programming | Test 1 | Question: 11

Question

What will be the outout of the following code?

#include <stdio.h>
int main()
{
    int a = 1, b = 2;
    int c = a++ || b++;
    printf("%d %d %d", a, b, c);
}

• A. 1 2 1
• B. 2 3 1
• C. 2 2 1
• D. 2 2 0

Answer 1

int a = 1, b = 2;
int c = a++ || b++;

2nd statement will run a++ and then there is logical OR operator so it will halt. (Short Circuit Rule for logical operators in C).

c = 1,
a = 2,
b = 2 (unaffected)

Comments

what are the short circuit rule in c plz explain

---

for logical AND, if the first operand is false (0), second is not evaluated. Similarly for logical OR, if the first operand is 1, the second one is not evaluated.

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a++ will be evaluated first or logical OR ..a++ is post increment so is it ???

int c = a || b;

a++ ; b++

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Option C is correct.

Output will be : 2 2 1

Answer 2

• || and && are logical OR and AND. (asked here)
   
• | and & are bitwise OR and AND. (not asked here)
   
• Check this: https://gateoverflow.in/44842/isro-2013-64

---

    int c = a++ || b++;

The logical OR would stop as soon as it finds any positive value ($\equiv$ 1). It stops right at a.

So, c = a

=> c = 1

 

Now, a++

=> a = 2

 

And b is never touched.

So, 2,2,1.

Option C